If the passing of five half-lives leaves 25.0 mg of a strontium-90 sample, how much was present in the beginning

initial amount = 800 mg

Explanation:

Given data:

Amount of strontium-90 after 5 half life = 25 mg

Total amount of sample = ?

Solution:

At 5th half life = 25 mg

At 4th half life = 25× 2= 50 mg

At 3rd half life = 50×2 = 100 mg

At 2nd half life = 100×2 = 200 mg

At 1st half life = 200×2 = 400 mg

when time 0 = 400×2 = 800 mg

Conformation:

initial amount = 800 mg

1st half life = 800/2 = 400 mg

2nd half life = 400/2 = 200 mg

3rd half life = 200/2 = 100 mg

4th half life = 100/2 = 50 mg

5th half life = 50 /2 = 25 mg

the correct answer is a

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890